Chapter 21 Electromagnetic Induction and Faraday’s Law

Chapter 21
Electromagnetic Induction and
Faraday’s Law
Induced EMF
Almost 200 years ago, Faraday looked for evidence that
a magnetic field would induce an electric current with
this apparatus:
He found no evidence when the current through the lefthand loop was steady, but did see a current induced in
the right-hand loop when the switch was turned on or
off.
In addition, a current will be induced in a wire loop if a
magnet is moved through the loop, but not when the
magnet is held steady.
Faraday’s Law of Induction;
Lenz’s Law
The induced emf in a wire loop is
proportional to the rate of change
of magnetic flux through the
loop.
Magnetic flux:
Unit of magnetic flux: weber,
Wb.
1 Wb = 1 T·m2
Faraday’s law of induction:
The emf induced in a circuit is equal to the rate of
change of magnetic flux through the circuit:
For coil with N number of turns:
The negative sign gives the direction of the induced emf.
Lenz’s Law:
An induced emf is always in a direction that opposed the
original change in flux that caused it.
Magnetic flux will change if the area of the loop changes,
and also with the angle between the loop and the field.
EMF Induced in a Moving Conductor
A uniform movable rod is
resting on a U shaped
conductor. If the rod is made
to move at speed 𝑣, it travels
a distance of ∆𝑥 = 𝑣∆𝑡 in
time ∆𝑡.
Hence by Faraday’s law:
∆Φ𝐵 𝐵Δ𝐴 𝐵𝑙𝑣Δ𝑡
𝜀=
=
=
Δ𝑡
Δ𝑡
Δ𝑡
= 𝐵𝑙𝑣
You can see using right hand rule that electrons on the rod
experiences a 𝐵𝑞𝑣 force such that they move to the top of
the rod. Thus the induced current is clockwise (to counter the
increasing flux.)
Changing Magnetic Flux
Produces an Electric Field
A changing magnetic flux induces an electric field; this
is a generalization of Faraday’s law. The electric field
will exist regardless of whether there are any conductors
around.
𝐸 =𝑣×𝐵
Where 𝑣 is the speed of the charged particle or object.
Electric Generators
A generator is the opposite of a motor—it transforms
mechanical energy into electrical energy. This is an ac
generator:
The axle is rotated by an
external force such as
falling water or steam.
Rotating axle causes
change in area of the
coil, thus changing flux.
This produces electric
current in the circuit
Transformers and Transmission of Power
A transformer consists of two
coils, either interwoven or linked
by an iron core. A changing emf in
one induces an emf in the other.
The ratio of the emfs is equal to
the ratio of the number of turns in
each coil.
P=IV
The figure aside shows a Step Up
transformer.
(21-6)
Inductance
Mutual inductance: a changing current in one coil will
induce a current in a second coil.
And vice versa; note that the constant M, known as the
mutual inductance, is the same:
Unit of inductance: the henry, H.
1 H = 1 V·s/A = 1 Ω·s
A transformer is an example
of mutual inductance.
A changing current in a coil
will also induce an emf in the
same coil (Lenz’s law):
Here, L is called the selfinductance
Problem 1:
An 18.5 cm diameter loop of wire is initially oriented
perpendicular to 1.5 T magnetic field. The loop is rotated so
that its plane is parallel to the field direction in 0.2s. What is
the average induced emf in the loop?
Solution:
∆Φ𝐵
𝐴𝐵∆𝑐𝑜𝑠𝜃
𝜀=−
=−
∆𝑡
∆𝑡
𝜋(0.0925)2 × 1.5 × (𝑐𝑜𝑠90 − 𝑐𝑜𝑠0)
=
= 0.20𝑉
0.2
Problem 2:
A circular loop in the plane of a paper lies in a 0.65T
magnetic field pointing into the paper. The loop’s diameter
changes from 20cm to 6 cm in 0.5 sec. What is (a) direction
of induced current? (b) average induced emf? (c) average
induced current if the coil resistance is 2.5 Ohms?
Solution:
(a)Flux is decreasing, so current produce will be to increase
the flux, i.e, the field should point into the paper. Hence
current should be clockwise.
(b) 𝜀 =
∆Φ𝐵
−
∆𝑡
(c) 𝐼 =
𝜀
𝑅
=
𝐵∆𝐴
−
∆𝑡
= 0.0015 𝐴
=0.0037V
Problem 3:
A transformer has 330 primary turns and 1240 secondary
turns. The input voltage is 120V and the output current is
15A. What are the output voltage and input current?
Solution:
(a)
(b)
𝐼𝑠
𝐼𝑃
=
𝑁𝑃
𝑁𝑆
𝑉𝑠
𝑁𝑠
=
𝑉𝑃 𝑁𝑃
1240
𝑉𝑠 =
× 120𝑉 = 450.9𝑉
330
1240
𝐼𝑃 =
× 15𝐴 = 56𝐴.
330
Problem 4:
Determine a formula for the self inductance 𝐿 of a long
tightly wrapped solenoid coil of length 𝑙 and cross sectional
area A, that contains N turns of wire.
Solution:
∆Φ𝐵
Δ𝐼
𝜀 = −𝑁
= −𝐿
Δ𝑡
Δ𝑡
ΔΦ𝐵
𝑆𝑜, 𝐿 = 𝑁
Δ𝐼
We know that for a solenoid, 𝐵 = 𝜇0 𝑁𝐼/𝑙, and Φ𝐵 = 𝐵𝐴.
𝜇0 𝑁𝐼𝐴
Thus, Φ𝐵 =
𝑙
𝜇 𝑁Δ𝐼.𝐴
So, ΔΦ𝐵 = 0
𝑙
ΔΦ𝐵
Hence, 𝐿 = 𝑁
Δ𝐼
=
𝜇0 𝑁2 𝐴
𝑙
LR Circuit
When the switch is closed,
current starts to flow, but
inductor opposes this flow
by producing induced emf
(Lenz’s law). Hence current
will increase less rapidly.
The current reaches steady
value when there is no more
emf in the inductor (I is no
longer changing.)
If the circuit is then shorted across the battery, the
current will gradually decay away.
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AC Circuits and Reactance
Resistors, capacitors, and inductors have different phase
relationships between current and voltage when placed
in an ac circuit.
The current through a resistor is in phase with the
voltage.
The current through an inductor lags the voltage by 90°.
The current reaches the peak (or trough) ¼ cycle after
the voltage does.
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In a capacitor, the current leads the voltage by 90°.
Reactance of a Circuit
Both the inductor and capacitor have an effective
resistance (ratio of voltage to current), called the
reactance.
Inductor:
Capacitor:
Note that both depend on frequency.
(21-11b)
(21-12b)
LRC Series AC Circuit
Analyzing the LRC series AC circuit is complicated, as
the voltages are not in phase—this means we cannot
simply add them. Furthermore, the reactances depend on
the frequency.
We calculate the voltage (and current) using what are
called phasors—these are vectors representing the
individual voltages.
Here, at t = 0, the current
and voltage are both at a
maximum. As time goes on,
the phasors will rotate
counterclockwise.
Some time t later, the phasors have rotated.
The voltages across each device are given by the
x-component of each, and the current by its
x-component. The current is the same throughout
the circuit.
We find from the ratio of voltage to current that the
effective resistance, called the impedance, of the circuit
is given by:
Phase angle is calculated as:
∅=
𝑡𝑎𝑛−1
𝑋𝐿 − 𝑋𝑐
𝑅
Resonance in AC Circuits
The rms current in an ac circuit is:
Clearly, Irms depends on the frequency.
We see that Irms will be a maximum when XC = XL; the
frequency at which this occurs is
This is called the resonant frequency.
Also, average Power dissipated:
𝑃𝑎𝑣 = 𝑉𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 𝑐𝑜𝑠∅
Problem 5:
An inductance coil operates at 240V and 60 Hz. It draws
12.2A. What is the coil’s inductance?
Solution:
𝑉 = 𝐼𝑋𝐿
𝑋𝐿 = 2𝜋𝑓𝐿
Hence,
𝑉
240
𝐿=
=
= 5.22 × 10−2 𝐻
2𝜋𝐹𝐼 2𝜋(60)(12.2)
Problem 6:
Determine the total impedance, phase angle, and rms
current in an LRC circuit connected to a 10kHz, 725V(rms)
source if L =28mH, R=8.7 k Ohm, and C=6250pF.
Solution:
(a)
(b) ∅ =
−1 𝑋𝐿 −𝑋𝑐
𝑡𝑎𝑛
𝑅
(c) 𝐼𝑟𝑚𝑠 =
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𝑉𝑟𝑚𝑠
𝑍
Problem 7:
The variable capacitor in the tuner of an AM radio has a
capacitance of 2800pF when the radio is tuned to a station at
580kHz. (a) What must be the capacitance for a station at
1600kHz? (b) What is the inductance?
Solution:
a). 𝑓 =
1
2𝜋
Hence,
1
𝐿𝐶
𝑓580
𝑓1600
=
𝐶1600
𝐶580
So,
580 2
1600
𝐶1600 = 367.9𝑝𝐹
b) 𝑓 =
1
2𝜋
So, 580 ×
1
𝐿𝐶
103
=
1
2𝜋
1
𝐿×2800×10−12
𝐿 = 2.7 × 10−5 𝐻
=
𝐶1600
2800×10−12